The half -life period of `._(84)Po^(210)` is 140 days.
In how many days `1 g` of this isotope is reduced to `0.25 g`?

Explanation

Original quantity of the isotope `(N_(0)) = 1 g`
Final quantity of the isotope N = 0.25 g
We know that, `N = ((1)/(2))^(n) N_(0)`
So, `(1)/(4) = ((1)/(2))^(n) xx 1`
or `((1)/(2))^(2) = ((1)/(2))^(n)`
or Time taken `T = n xx t_(1//2) = 2 xx 140 = 280` days